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andreev551 [17]
3 years ago
8

a driver dove to a location 6 3/5 meters below sea level he then dove to a second location 8 1/5 meters below sea level how many

meters are there between the two locatiojn
Mathematics
1 answer:
Mariana [72]3 years ago
8 0
Answer: 1 3/5 meter below sea level

In this case, a diver goes into two different location expressed with "meters below sea level" unit. To know the distance between the two location, then you need to subtract it. 
Distance = 2nd location - 1st location 
Distance = 8 1/5 meter below sea level - 6 3/5 meter below sea level= 1 3/5 meter below sea level
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If f(x) = x2 – 2x and g(x) = 6x + 4, for which value of x does (f + g)(x) = 0?
Hatshy [7]

Answer:

x = - 2

Step-by-step explanation:

(f + g)(x)

= f(x) + g(x)

= x² - 2x + 6x + 4

= x² + 4x + 4

Equating to zero

x² + 4x + 4 = 0

(x + 2)² = 0

x + 2 = 0

x = - 2

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If (9 to the 2nd power)p = 9 to the 8th power , what is the value of p?
Novay_Z [31]

(9^2)^p=9^8\\\\9^{2p}=9^8\\\\2p=8\\\\p=4

3 0
3 years ago
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What is 1 1/12+4 5/12 pls
lilavasa [31]

Answer:

5 1/3 or 5.333 or 16/3

Step-by-step explanation:

Yespp

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3 years ago
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Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.
Cloud [144]

If F_Y(y) is the cumulative distribution function for Y, then

F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)

Then the probability density function for Y is f_Y(y)={F_Y}'(y):

f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}

The nth moment of Y is

E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy

Let u=\ln y, so that \mathrm du=\frac{\mathrm dy}y and y^n=e^{nu}:

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du

Complete the square in the exponent:

nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du

But \frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2} is exactly the PDF of a normal distribution with mean n and variance 1; in other words, the 0th moment of a random variable U\sim N(n,1):

E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1

so we end up with

E[Y^n]=e^{\frac12n^2}

3 0
3 years ago
Find the midpoint of the segment between the points (17,−11) and (−14,−16)
Viefleur [7K]

Answer:

(1.5, -13.5)

Step-by-step explanation:

Midpoint Formula: (\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )

Simply plug in our coordinates into the formula:

x = (17 - 14)/2

x = 3/2

y = (-11 - 16)/2

y = -27/2

4 0
3 years ago
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